There are 2n different balls so that each ball numbered from 1 to 2n. There are 2n different boxes numbered from 1 to 2n. How many ways are there so that there is a single even numbered ball in an even numbered box(the box and the ball can have different numbers - example ball number - 4, box number 6. they just have to both be even).
My solution for some reason isnt true - Pick an even numbered box - n choices, to which we will place an even numbered ball - n choices. Now to make sure the rest of even numbered boxes get uneven numbered balls - we have n uneven balls so we pick a random even box and have n choices, then n-1,..., and we get n!/2 choices to stuff all n-1 even boxes with uneven numbered balls. Now we have 1 uneven ball left and n-1 even balls, so n balls in total. We can randomly distribute them. so n!.
So my answer is (nn(n!)^2)/2=n^3(n!)/2
Where did I go wrong?
The correct answer is n^4*((n-1)!)^2